Answer:
26.5 g
Explanation:
First we convert 100.0 mL to L:
- 100.0 mL / 1000 = 0.100 L
Now we <u>calculate how many moles of sodium carbonate are needed</u>, using the <em>definition of molarity</em>:
- Molarity = moles / liters
- moles = molarity * liters
- 2.5 M * 0.100 L = 0.25 mol
Finally we <u>convert 0.25 moles of sodium carbonate into grams</u>, using its <em>molar mass</em>:
- 0.25 mol * 106 g/mol = 26.5 g
To calculate the mass of Fe formed in a) we get first the limiting reactant between Fe2O3 and CO. Given the masses, the ratio of Fe2O3 is 1.33 while that of CO is 1.67. Hence the limiting reagent is Fe2O3. The mass of Fe formed is 148.98 grams. In b, the needed CO is only 112.04 grams. Hence, the excess is 27. 96 grams.
A triple bonded carbon is called an alkyne. (with a y, e for double bond, a for single bond)
Answer:
[Br₂] = 1.25M
Explanation:
2NO (g) + Br₂ (g) ⇄ 2NOBr (g)
Eq 0.80M ? 0.80M
That's the situation told, in the statement.
Let's make the expression for Kc
Kc = [NOBr]² / [Br₂] . [NO]²
Kc = 0.80² / [Br₂] . [0.80]²
0.80 = 1 / [Br₂]
[Br₂] = 1 / 0.80 → 1.25
